Given that \(\displaystyle{A}_{{{2}}}\) is the set of all multiples of 2 except for 2. In this set all the even numbers except 2 are covered.

Similarly, \(\displaystyle{A}_{{{n}}}\) is defined such that it is the set of all multiples of n except for n.

Now consider a composite number m.

Then \(\displaystyle{m}={a}{b}\) with \(\displaystyle{\left({a},{b}\right)}={1}\)

Then the number m belongs to the set \(\displaystyle{A}_{{{a}}}\) and \(\displaystyle{A}_{{{b}}}\)

Hence it belong to the set \(\displaystyle{A}_{{{1}}}\cup{A}_{{{2}}}\cup{A}_{{{3}}}\cup\cdots\) And thus it cannot belong the set \(\displaystyle\overline{{{A}_{{{1}}}\cup{A}_{{{2}}}\cup{A}_{{{3}}}\cup\cdots}}\)

Step 2

On the other hand, consider a prime number p.

Then it cannot be written as a multiple of any element other than p and 1.

By the definition of the set \(\displaystyle{A}_{{{p}}}\), p does not belong to \(\displaystyle{A}_{{{p}}}\)

which gives that p belongs to \(\displaystyle\overline{{{A}_{{{1}}}\cup{A}_{{{2}}}\cup{A}_{{{3}}}\cup\cdots}}\)

Finally, the set of all odd primes will constitute the set \(\displaystyle\overline{{{A}_{{{1}}}\cup{A}_{{{2}}}\cup{A}_{{{3}}}\cup\cdots}}\)